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it is given that dy = 3y. You can take it from there.

After you get that, do you integrate both sides?

the problem is not clear. does y change at a rate exactly three times its previous value?

I would assume so because it does say values...does that help any?

well, if y(1) = 4e, then y(2) = 3 times 4e and y(3) = 3 times 3times 4e and so on

hmm so that would be 105e? That doesn't feel right, but I mean the steps make sense.

no wait. that is wrong. scrap that.

The second one or both?

the second one.

So, the first one was indeed right?

I think so. Is this problem in application of derivatives or integrals or something?

they have given y(1) = 4e. what is the 1? is it y at 1 second? is y a function of time?

I can't see it being a function of time...There is not information relating the problem to time.

where rate is given, it is generally a function of time.

Oh, I completely overlooked the rate of it all.

so dy(t)/dt = 3y(t-1) ?

Yes, so is that just the original function of y or is this y at 1?

no, it is the general function of y.

because it is given that it changes at 3 times its values.

Ahhhh haaa!

yes, a function can depend on its previous values. you can put an f in the function.

no no, t = 1.

you are welcome.